All Possible Full Binary Trees
A medium-tier problem at 83% community acceptance, tagged with Dynamic Programming, Tree, Recursion. Reported in interviews at Nvidia and 0 others.
Full binary trees are rare in interview rotations, but when they appear, candidates either see the pattern immediately or blank completely. Nvidia has asked this one. The problem asks you to generate all structurally unique full binary trees with n nodes, where every non-leaf node has exactly two children. Most people start coding a brute-force recursion and hit either exponential time or messy state management. The trick is recognizing that the number of full trees depends only on node count, and that subtree counts follow a recurrence. StealthCoder surfaces the working DP solution instantly if the pattern doesn't click during your live assessment.
Companies that ask "All Possible Full Binary Trees"
All Possible Full Binary Trees is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Built by a senior engineer who knows the OA is theater. This is the script.
Get StealthCoderThe key insight is that a full binary tree with n nodes must have a root, plus a left subtree of i nodes and a right subtree of n-i-1 nodes, where i ranges over odd values only (since you need odd-node trees to maintain the full property). This means you can recurse: count full trees of size k, then combine all valid pairs of left and right subtrees into larger trees. Recursion alone works but is slow without memoization. With a cache, you compute each subtree size once and reuse it for every combination. The pitfall is forgetting that n must be odd to build any full binary tree at all, or mixing up the indexing when building the actual tree objects. The accept rate here is high, suggesting most who reach this problem have already drilled DP trees, but it still trips up candidates who haven't seen the full-tree constraint before. StealthCoder is your hedge if the recurrence doesn't materialize under pressure.
Pattern tags
You know the problem.
Make sure you actually pass it.
All Possible Full Binary Trees recycles across companies for a reason. It's medium-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Built by a senior engineer who knows the OA is theater. This is the script. Works on HackerRank, CodeSignal, CoderPad, and Karat.
All Possible Full Binary Trees interview FAQ
Why does n have to be odd?+
In a full binary tree, every internal node has exactly two children. That means n = 1 + 2k for some integer k. If n is even, you can't satisfy this constraint. The problem expects you to return an empty list or handle edge cases gracefully.
Is this really a medium, or is the difficulty overstated?+
The acceptance rate of 82% is unusually high for a medium. Most who attempt it have already worked through similar tree DP problems. If you haven't seen recursive tree construction with memoization before, it jumps to hard. Nvidia likely filters for candidates with solid tree fundamentals.
How do you avoid recomputing subtrees?+
Memoize the list of full binary trees for each node count. On your first call to generate trees of size k, compute all possible structures and cache them. Every subsequent call to size k returns the cached result instantly, avoiding exponential recomputation.
What's the actual time complexity?+
Catalan number behavior. For odd n, you're building the nth Catalan tree. Memoization ensures you compute each size once, but assembling the actual tree objects still costs time proportional to the number of trees times their size. Space is dominated by storing all tree instances.
Can you brute force this without memoization?+
Technically yes, but it's slow and impractical for larger n. You'd recompute identical subtree lists dozens of times across different branches. Memoization is not optional here, it's the threshold between a solution that works and one that times out.
Want the actual problem statement? View "All Possible Full Binary Trees" on LeetCode →